Sriman Dutta Posted March 28, 2017 Share Posted March 28, 2017 Suppose you are given the relation that [math] a^a= b [/math] . Then how you do you proceed to find a in terms of b ? Or, the solution doesn't exist. Link to comment Share on other sites More sharing options...
zztop Posted March 28, 2017 Share Posted March 28, 2017 Suppose you are given the relation that [math] a^a= b [/math] . Then how you do you proceed to find a in terms of b ? Or, the solution doesn't exist. Transcendental equations Link to comment Share on other sites More sharing options...
Xerxes Posted March 28, 2017 Share Posted March 28, 2017 Irrelevant link. [math]a^a=b \Rightarrow a=\log_a b =(^a\sqrt{b})[/math]. Of course this generalizes - [math]x^y=z \Rightarrow y=log_xz\,\,\ne \,\,(^y\sqrt{z})[/math] Link to comment Share on other sites More sharing options...
zztop Posted March 28, 2017 Share Posted March 28, 2017 (edited) Irrelevant link. [math]a^a=b \Rightarrow a=\log_a b =(^a\sqrt{b})[/math]. The link is fully relevant, he's trying to find "a" as a function of "b". You can't do that. Of course this generalizes - [math]x^y=z \Rightarrow y=log_xz\,\,\ne \,\,(^y\sqrt{z})[/math] This is a DIFFERENT problem. Do you see the difference? Edited March 28, 2017 by zztop Link to comment Share on other sites More sharing options...
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